Brian Bi
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Return to table of contents for Brian's unofficial solutions to Artin's Algebra

Section 2.6. Isomorphisms

Exercise 2.6.2 The homomorphism \(\varphi\) is completely determined by the image of 1, since 1 generates the domain. If \(\varphi(1) = 0\), then \(\varphi\) is neither surjective nor injective. If \(\varphi(1) = \pm 1\), then \(\varphi(n) = \pm n\), and \(\varphi\) is both surjective and injective (hence, an isomorphism). If \(\varphi(1) = k\) where \(k \neq \pm 1\), then \(\varphi(n) = kn\) and \(\varphi\) is injective, but not surjective as 1 is not in the image.

Exercise 2.6.4 \(ba = b(ab)b^{-1}\), the conjugate of \(ab\) by \(b\).

Exercise 2.6.5 The eigenvalues of \(B\) are 2 and 3. Since \(B\) is a 2×2 matrix, the multiplicity of each eigenvalue is exactly 1, therefore \(A\) is the Jordan form of \(B\) (up to the order of diagonal elements). Therefore \(A\) and \(B\) are similar, which is the same as saying that they are conjugate in \(GL_2(\mathbb{R})\).

Exercise 2.6.6 Denote the two given matrices by \(A, B\). They are conjugate by the matrix \(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\), so they are conjugates in \(GL_2(\mathbb{R})\). In general, \(M = \begin{pmatrix} p & q \\ r & s \end{pmatrix}\), then solving the equation \(BM = MA\) yields the constraints \(p = 0, q = r\), implying \(\det M = -q^2\), which is not possible when \(\det M = 1\) and \(q\) is real. Therefore \(A\) and \(B\) are not conjugate elements in \(SL_2(\mathbb{R})\).

Exercise 2.6.7 Since \(1 \in H\), it follows that \(g(1)g^{-1} = 1\) is in \(gHg^{-1}\). For all \(k \in gHg^{-1}\), there eixsts some \(h \in H\) with \(k = ghg^{-1}\), whence \(k^{-1} = gh^{-1}g^{-1}\), which lies in \(gHg^{-1}\) since \(h^{-1} \in H\). Finally, for all \(k_1, k_2 \in gHg^{-1}\), there exist some \(h_1, h_2 \in H\) with \(k_1 = gh_1 g^{-1}, k_2 = gh_2 g^{-1}\), implying \(k_1 k_2 = gh_1 g^{-1} g h_2g^{-1} = g(h_1 h_2)g^{-1} \in gHg^{-1}\) since \(h_1 h_2 \in H\). Therefore \(gHg^{-1}\) is a subgroup of \(G\).

Exercise 2.6.8 For all \(A, B \in GL_n(\mathbb{R})\), we have \(\varphi(AB) = ((AB)^t)^{-1} = (B^t A^t)^{-1} = (A^t)^{-1} (B^t)^{-1} = \varphi(A)\varphi(B)\). Therefore \(\varphi\) is an endomorphism of \(GL_n(\mathbb{R})\). It is easy to see that the kernel of \(\varphi\) is trivial, making \(\varphi\) an automorphism.

Exercise 2.6.9 We claim that the map \(\varphi\) from \(G\) to \(G^\circ\) given by \(\varphi(g) = g^{-1}\) is an isomorphism. (We established in Exercise 2.2.6 that \(G^\circ\) is indeed a group.) Let \(a, b \in G\). Then \(\varphi(ab) = (ab)^{-1} = b^{-1} a^{-1} = a^{-1} * b^{-1} = \varphi(a) * \varphi(b)\). Therefore \(\varphi\) is a homomorphism. Its kernel is trivial, so it is an isomorphism.

Exercise 2.6.10

  1. Let \(\varphi : C_{10} \to C_{10}\) be an automorphism. Since \(C_{10}\) is cyclic, the homomorphism \(\varphi\) is entirely determined by \(\varphi(1)\), since \(\varphi(2) = \varphi(1) + \varphi(1)\) and so on; in general, every valid endomorphism must take the form \(\varphi(n) = kn\). For such \(\varphi\), we have \(\varphi(a+b) = k(a+b) = ka + kb = \varphi(a) + \varphi(b)\) for all \(a, b\), so \(\varphi\) preserves the group operation. Therefore the set of such \(\varphi\) is exactly the set of endomorphisms of \(C_{10}\). Note that \(\varphi\) can only take on values that are divisible by \(\gcd(k, 10)\), so \(k\) must be congruent to 1, 3, 7, or 9 in order for \(\varphi\) to be an automorphism. As proven in the solution to Exercise 2.4.6, the elements 1, 3, 7, and 9 generate \(C_{10}\), so \(\varphi(n) = kn\) for \(k \in \{1, 3, 7, 9\}\) maps a generator to a generator, implying that it is surjective, and hence, in this case, also injective, therefore an automorphism. Therefore, the automorphisms of \(C_{10}\) are \(n \mapsto n, n \mapsto 3n, n \mapsto 7n\), and \(n \mapsto 9n\).
  2. An endomorphism \(\varphi : S_3 \to S_3\) is completely determined by the images of the generators \(x, y\); there will be exactly one valid extension to all of \(S_3\) provided that \(\varphi(x)\) and \(\varphi(y)\) are chosen so that \begin{equation} \varphi(y)\varphi(x) = \varphi(x)^2 \varphi(y) \label{eqn:e2} \end{equation} that is, \(\varphi\) also preserves the defining relation of \(S_3\). If \(\varphi\) is additionally an automorphism, it must preserve the orders of elements. Referring to the table from Exercise 2.2.1, we see that \(x\) and \(x^2\) have order 3 while \(x^j y\) has order 2 for \(j \in \{0, 1, 2\}\). Therefore each valid automorphism must take the form \(\varphi_{ij}\) which satisfies \(\varphi_{ij}(x) = x^i, \varphi_{ij}(y) = x^j y\), with \(i \in \{1, 2\}\) and \(j \in \{0, 1, 2\}\). Then \begin{align} \varphi(y)\varphi(x) &= x^j y x^i \label{eqn:e1} \\ \varphi(x)^2\varphi(y) &= x^{2i} x^j y \end{align} Since \(yx = x^2 y\), we can move the \(y\) factor \(i\) positions to the right in \((\ref{eqn:e1})\) to obtain \(\varphi(y)\varphi(x) = x^{2i+j} y\); therefore \((\ref{eqn:e2})\) is always satisfied, so the 6 different possible maps \(\varphi_{ij}\) are all valid endomorphisms of \(S_3\).

    The general formula for \(\varphi_{ij}\) is, of course, \begin{equation} \varphi_{ij}(x^m y^n) = \varphi_{ij}(x^m) \varphi_{ij}(y^n) = x^{im} (x^j y)^n \end{equation} Suppose this is equal to 1. Now \(x^j y\) has order 2 so a factor of \(y\) will be left in the result unless \(n\) is even. In the case where \(n\) is even, \(\varphi_{ij}(x^m y^n) = x^{im}\), so \(im\) must be divisible by 3, and since \(i\) is either 1 or 2, this implies \(m\) is divisible by 3. Therefore, if \(\varphi_{ij}(g) = 1\) then \(g = 1\), that is, each of the 6 \(\varphi_{ij}\)'s is an automorphism of \(S_3\), and there are no other automorphisms.

Exercise 2.6.11 Since \(\{1, a\}\) is a normal subgroup, for all \(g \in G\), \(gag^{-1} \in \{1, a\}\). But \(gag^{-1}\) cannot be equal to 1 unless \(a = 1\), therefore \(gag^{-1} = a\) for all \(g \in G\), that is, \(a\) is central.